Use the following balanced equation,
CaBr2(aq) + H2SO4(aq) ? CaSO4(s) + 2HBr(g)
If 10.32 g of HBr are generated, what volume of 0.100 M H2SO4 is needed to produce this much HBr?
lewis structure for SbF5?
The Correct Answer and Explanation is :
Let’s start with the first problem:
Given balanced equation:
[
\text{CaBr}_2(aq) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{CaSO}_4(s) + 2\text{HBr}(g)
]
Step 1: Convert Grams of HBr to Moles
The molar mass of HBr is calculated as:
[
\text{HBr} = 1.008 + 79.904 = 80.912 \text{ g/mol}
]
Given:
[
10.32 \text{ g HBr} \times \frac{1 \text{ mol HBr}}{80.912 \text{ g HBr}} = 0.1275 \text{ mol HBr}
]
Step 2: Determine Moles of H₂SO₄ Needed
From the balanced equation, 2 moles of HBr are produced per 1 mole of H₂SO₄, so:
[
0.1275 \text{ mol HBr} \times \frac{1 \text{ mol H}_2\text{SO}_4}{2 \text{ mol HBr}} = 0.06375 \text{ mol H}_2\text{SO}_4
]
Step 3: Calculate Volume of H₂SO₄ Solution
Given that the concentration of H₂SO₄ is 0.100 M:
[
\text{Volume} = \frac{\text{moles of H}_2\text{SO}_4}{\text{Molarity}} = \frac{0.06375}{0.100} = 0.6375 \text{ L} = 637.5 \text{ mL}
]
Final Answer:
To produce 10.32 g of HBr, 637.5 mL of 0.100 M H₂SO₄ is required.
2. Lewis Structure of SbF₅
To determine the Lewis structure for SbF₅ (Antimony Pentafluoride):
- Count Valence Electrons:
- Sb (Antimony) is in Group 15 → 5 valence electrons.
- F (Fluorine) is in Group 17 → 7 valence electrons.
- Total for SbF₅ = 5 + (5 × 7) = 40 valence electrons.
- Draw the Skeleton Structure:
- Place Sb in the center (less electronegative).
- Attach five fluorine atoms around it.
- Distribute Electrons:
- Each F needs 8 electrons (octet rule).
- All five fluorines are single bonded to Sb, using 10 bonding electrons (5 bonds × 2 electrons).
- Remaining 30 electrons go to fluorines as lone pairs (6 electrons per F).
- Check Formal Charges:
- Sb has five single bonds (0 lone pairs).
- Each F has three lone pairs and one bond.
- All atoms have formal charges of zero, making it stable.
Since SbF₅ follows the AX₅ molecular geometry (VSEPR theory), it has a trigonal bipyramidal shape.
I will generate the Lewis Structure image now.
Here is the Lewis structure diagram for SbF₅. You can see that the central Sb atom is bonded to five fluorine atoms in a trigonal bipyramidal arrangement, with each fluorine obeying the octet rule. Let me know if you need any further clarification!